More precisely, if $X \rightarrow \mathbb{A} ^n$ is a flat map, does there exist a map $Y \rightarrow X$ such that, for every $p \in \mathbb{A} ^n$, the fiber map $Y_p \rightarrow X_p$ is a resolution of singularities? Can one require, moreover, that the map $Y \rightarrow \mathbb{A} ^n$ is smooth?

2 Answers
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I assume you want $Y \to X$ to be proper. The answer is a definite no, in general. For example, take a polynomial $f: \mathbb A^2 \to \mathbb A^1$; such a $Y$ would have to be finite over $\mathbb A^2$, and birational, so $Y = \mathbb A^2$. There are lots of counterexamples in higher dimension too: for example, it follows from the purity theorem that you usually can't have a simultaneous resolution when $X$ is smooth. Thus, for example, in the very simple example $f\colon \mathbb A^3 \to \mathbb A^1$, $f(x, y, z) = x^2 + yz$, in which the only singular fiber is over the origin, and it has the simplest kind of surface singularity, of type $A_1$, you don't have a simultaneous resolution.

There are some non-trivial results, but they require the base to be 1-dimensional, and they require a base change on the base to get the resolution. For example, in the example above if one makes a base change $t \mapsto t^2$ on the base and normalizes, one has a simultaneous resolution. This is particular case of a theorem of Brieskorn: see for example M. Artin, Algebraic construction of Brieskorn's resolutions, Journal of Algebra 29 (1974). This is only possible in very particular cases, though.